Stoichiometry

Question 1

Chromatography

Answer 1

The Process where a mobile phase (Which Contains the mixture to be analysed) moves over a stationary phase. As this happens, a process of adsorption and desorption leads to some components of the mixture moving faster over the stationary phase than others

Question 2

Adsorption

Answer 2

The process by which molecules stick to the surface by other particles. It commonly involves liquid or gas molecules sticking to the surface of solid particles.

Question 3

HPLC

Answer 3

High performance liquid chromatography is a method for measuring the concentration of organic compounds in a sample. It allows organic molecules at low concentrations to be separated from a mixture. These components can then be quantitatively.

Question 4

Paper and Thin layer Chromatography

Answer 4

Submerging the edge of a paper into a solvent and watching the sample rise up the paper and undergo a continual process of adsorption and desorption. It aims to seperate out the components in the mixture and identify the presence of a component

Question 5

Identifying sample components in thin layer chromatrography.

Answer 5

Rf (Retardation factor). This value describes how far the component has travelled up the stationery phase.
rf= 0 Means that component remaind at stationery phase
rf = 1 Means that component is at solvent front and showed no affinity for the stationery phase
rf = inbetween means it showed affinity for both stationary and mobile phase.

Question 6

Equation to find length (Paper Chromo)

Answer 6

Distance travelled divided by total length from origion to solvent end.

Question 7

A Chromatogram provides information about

Answer 7

The identity each component of a sample via the rentention time (Qualitative Anaysis).
The Concentration of each component of a sample via peak area (Quantitiatve Analysis)

Question 8

Order of Molecules From most to least Polar

Answer 8

Most Polar

• Carboxylic acid (Hydrogen, dispersion , dipole, 3 Polar bonds)
• Alcohol (Hyrdogen, Dipole, Dispersion, 2 Polar Bonds)
• Ester (Dipole, Dispersion)
• Alkane (Weak Dispersion)

Question 9

Concentration of Cl in 250ml (V) of solution 52.2g (Mass) of Aluminium chloride (AlCl3) is

Answer 9

4.76M

n(AlCl3) = 52.9 divided by 133.5  = 0.396
n(Cl) = 3 x 0.396 = 1.18 mol 
C = 1.18 divided by 0.25 = 4.76M

Question 10

Source of Error Titration (Rinsing water left in Burette)

Answer 10

Burette (Underestimated)
Titration Flask (Overestimated)

Question 11

Source of Error Titration (Rinsing Water left in Pipette)

Answer 11

Burette (Overestimated)
Titration Flask (Underestimated)

Question 12

Source of Error Titration ( Indicator Chosen Changes Colour too Soon)

Answer 12

Burette (Overestimated)
Titration Flask (Underestimated)

Question 13

Source of Error Titration (Water in Titration Flask)

Answer 13

Burette (No effect)
Titration Flask (No effect)

Question 14

Source of Error Gravimetric (Insufficient Precipitating agent was added)

Answer 14

Underestimated

Question 15

Source of Error Gravimetric (Precipitate not fully dried before weighing)

Answer 15

Overestimated

Question 16

Source of Error Gravimetric (Precipitate not washed prior to drying and weighing)

Answer 16

Overestimated

Question 17

Source of Error Gravimetric (Other Ions were present in the sample that reacted with the precipitating agent)

Answer 17

Overestimated

Question 18

Source of Error Gravimetric (Using too much water for the initial dissolving)

Answer 18

No effect

Question 19

Source of Error Gravimetric (Forming a precip thats too soluble)

Answer 19

Underestimated

Question 20

Limestone contains 95% CaCO3 by mass. The reaction is
CaCO3 (s) + 2HCL (aq) -> CaCl2 (aq) + CO2 (g) + H2O (g)

0.750 of Co2 was evolved. The amount in mol of HCL acid reacting was

Answer 20

0.750 divided by 44 = 0.017 x 2

= 3.41 x 10 to the power of negative 2

Question 21

Limestone contains 95% CaCO3 by mass. The reaction is
CaCO3 (s) + 2HCL (aq) -> CaCl2 (aq) + CO2 (g) + H2O (g)

The mass of lime stone added to the acid was

Answer 21

0. 17x 100.1 (Caco3) = 1.1
1. 7 x 100 Divided by 95 =
2. 80g.

Question 22

What will least likely alter the retention time of a substance passing through the column in a HPLC analysis

Answer 22

Diluting the sample being analysed (Will change peak but not retention time)

Question 23

Things that will alter retention time in a HPLC analysis

Answer 23

• Replacing one solvent with another
• Increasing the temp of the column
• Decreasing the rate of flow of the mobile phase

Question 24

How can Chromium be selectively analysed without interference from other metals

Answer 24

Because AAS has a hollow cathode lamp which is made from elements under investigation

Question 25

How can a particular wave length used for analysis is chosen in UV- visible spectrometry

Answer 25

We will choose length that is highly absorbed in the component. We will also choose length that is exclusively absorbed. This assures accurate results.

Question 26

The amount of NaCl in water was determined by Gravimetric Analysis. A 25.0ml of water was mixed with 100ml of distilled water, stirred and mixture filtered. The Volume of filtrate was made up to 250ml. A 20ml aliquot of solution was pipetted into a conical flask and an excess of AgNo3 solution was added. The Precip weighed 0.150g AgNo3+Nacl -> AgCl+ NaNO3 Calc amount in mol of precip

Answer 26

0.150 divided by 143.4 = 1.046x10^-3 mol

Question 27

The amount of NaCl in water was determined by Gravimetric Analysis. A 25.0ml of water was mixed with 100ml of distilled water, stirred and mixture filtered. The Volume of filtrate was made up to 250ml. A 20ml aliquot of solution was pipetted into a conical flask and an excess of AgNo3 solution was added. The Precip weighed 0.150g AgNo3+Nacl -> AgCl+ NaNO3 Calc amount in mol of NaCl in the 250ml solution

Answer 27

1:1 n (Nacl) =n (AgCl) = 1.046x10^-3 250 divided by 20 x 1.046x 10^-3 = 0.0131 mol

Question 28

The amount of NaCl in water was determined by Gravimetric Analysis. A 25.0ml of water was mixed with 100ml of distilled water, stirred and mixture filtered. The Volume of filtrate was made up to 250ml. A 20ml aliquot of solution was pipetted into a conical flask and an excess of AgNo3 solution was added. The Precip weighed 0.150g AgNo3+Nacl -> AgCl+ NaNO3 Calc Concentration in %m/v of sodium chloride in water

Answer 28

m (Nacl) = nx M (0.0131) x(58.5) = 0.766g in 25ml of water 0. 766 divided by 25 X 100 = 3. 06%

Question 29

The amount of NaCl in water was determined by Gravimetric Analysis. A 25.0ml of water was mixed with 100ml of distilled water, stirred and mixture filtered. The Volume of filtrate was made up to 250ml. A 20ml aliquot of solution was pipetted into a conical flask and an excess of AgNo3 solution was added. The Precip weighed 0.150g AgNo3 (aq) +Nacl (aq) -> AgCl (s) + NaNO3 (aq) State one property that Agcl must have to be used in the gravimetric analysis

Answer 29

- Low Solubility but must be stable when heated | - Have a Known formula

Question 30

Titration. (Determining Ammonia Content in Window Cleaner) 9.971g sample of cleaner was transformed to a 250ml standard flask and the volume made up to the mark with distilled water. A 20 ml aliquot of diluted window cleaner was pipetted into a conical flask and titrated with 0.110M HCl solution. The average Titre was 23.5 ml . HCl (aq) + NH3 (aq) -> NH4Cl (aq) Determine amount in mol of HCl used in the titration

Answer 30

n(Hcl) = Cv (0.110) (0.0235) = 2.59x 10^-3 mol

Question 31

Titration. (Determining Ammonia Content in Window Cleaner) 9.971g sample of cleaner was transformed to a 250ml standard flask and the volume made up to the mark with distilled water. A 20 ml aliquot of diluted window cleaner was pipetted into a conical flask and titrated with 0.110M HCl solution. The average Titre was 23.5 ml . HCl (aq) + NH3 (aq) -> NH4Cl (aq) Determine amount in mol of Ammonia in the 9.971g sample

Answer 31

1:1 n(NH3) = n (HCl) = 2.59 x10 ^ -3 n (NH3) in 250ml flask= 250 divided by 20 x 2.59x 10 ^-3 = 3.23x 10^-2

Question 32

Titration. (Determining Ammonia Content in Window Cleaner) 9.971g sample of cleaner was transformed to a 250ml standard flask and the volume made up to the mark with distilled water. A 20 ml aliquot of diluted window cleaner was pipetted into a conical flask and titrated with 0.110M HCl solution. The average Titre was 23.5 ml . HCl (aq) + NH3 (aq) -> NH4Cl (aq) Determine mass of Ammonia in the 9.971g sample

Answer 32

m (NH3) = nx M = 3.23 x 10^-2 x 17 =0.546g

Question 33

Titration. (Determining Ammonia Content in Window Cleaner) 9.971g sample of cleaner was transformed to a 250ml standard flask and the volume made up to the mark with distilled water. A 20 ml aliquot of diluted window cleaner was pipetted into a conical flask and titrated with 0.110M HCl solution. The average Titre was 23.5 ml . HCl (aq) + NH3 (aq) -> NH4Cl (aq) Determine Percentage by mass (%m/m) of ammonia in cleaner

Answer 33

0.549 Divided by 9.971 x 100 = 5.51 %

Question 34

During Titration the burette was incorrectly given a final rinse with distilled water. Would this lead to a higher or lower than expected result?

Answer 34

Higher than true value because the acid has been diluted so you will need more of it on the base